# QM - calculating expectation value for velocity of an electron

How do we calculate the expectation value for speed? I have heard that we must first calculate the expectaion value for kinetic energy. Someone please explain a bit what options do we have.

• Well i asked this because I cant find an anwser in my book which only describes operators of kinetic energy, potential energy, momentum, position and position square... If i write expectation value for speed, Google returns mostly these same operators... – 71GA Aug 7 '13 at 8:56
• Never mind I found it in this PDF - page 32 - Now give me some downvotes for being lazy :D – 71GA Aug 7 '13 at 8:59
• You can't derive a vector ($\vec v$) from its square ($T\sim v^2$) – Ruslan Aug 7 '13 at 9:10
• Well if you want just magnitude of velocity, then it's possible. – Ruslan Aug 7 '13 at 9:15
• I think this still isn't quite good (compare $\sqrt{\left<v^2\right>}$ with $\left<|v|\right>$ for some random set of $v$), you'd better compute it from velocity operator not from kinetic energy. – Ruslan Aug 7 '13 at 10:01

Calculating the energy eigenvalue, will give you $\langle v^2 \rangle$. This is how it's done:

$$\langle v^2 \rangle = \frac{2}{m}\langle T \rangle=\frac{2}{m}\langle \psi|\hat T|\psi\rangle=\frac{2}{m}\int\psi^*(x)\hat T \psi(x)dx$$

Where you should write $T$(the kinematic energy) as an operator. This can be done by writing it as a function of $x$ and $p$, and then replacing $p$ with its operator.

However this is not what we call the expectation value of speed. To calculate the expectation value of speed, we calculate the expectation value of its momentum:

$$\langle v \rangle = \frac{\langle p\rangle}{m}=\frac{1}{m}\int \psi^*(x)p\psi(x)dx$$

Which can be calculated either by transforming $\psi$ to the momentum space, or replacing $p$ with its operator $-\mathfrak i \hbar\frac{\partial}{\partial x}$.

An important thing to note is: $\langle v^2 \rangle \ne \langle v\rangle^2$ in general. E.g. consider the symmetric harmonic potential where $\langle v \rangle=0$ but $\langle v^2 \rangle > 0$. The difference is called variance $\sigma_v^2=\langle v^2 \rangle-\langle v \rangle^2$.

• Regarding your 1st paragraph. Why did you use the energy $\langle E \rangle$ and an operator $\hat{H}$ instead of $\langle E_k \rangle$ and an operator $\langle \hat{T}\rangle$? – 71GA Aug 7 '13 at 11:41
• Oops, that's a mistake. – Ali Aug 7 '13 at 13:39